Math 516 Fall 2006 Radford Written Homework # 5 Solution 12/12/06 Throughout R is a ring with unity. Comment: It will become apparent that the module properties 0 m = 0, (r m) = ( r) m, and (r r ) m = r m r m are vital details in some problems. 1. (20 total) Let M be an (additive) abelian group and End(M) be the set of group homomorphisms f : M M. (a) (12) Show End(M) is a ring with unity, where (f+g)(m) = f(m)+g(m) and (fg)(m) = f(g(m)) for all f, g End(M) and m M. Solution: This is rather tedious, but not so unusual as a basic algebra exercise. The trick is to identify all of the things, large and small, which need to be verified. We know that the composition of group homomorphisms is a group homomorphism. Thus End (M) is closed under function composition. Moreover End (M) is a monoid since composition is an associative operation and the identity map I M of M is a group homomorphism. Let f, g, h End (M). The sum f + g End (M) since M is abelian as (f + g)(m + n) = f(m + n) + g(m + n) = f(m) + f(n) + g(m) + g(n) = f(m) + g(m) + f(n) + g(n) = (f + g)(m) + (f + g)(n) for all m, n M. Thus End (M) is closed under function addition. 1
Addition is commutative since f + g = g + f as (f + g)(m) = f(m) + g(m) = g(m) + f(m) = (g + f)(m) for all m M. In a similar manner one shows that addition is associative which boils down to ((f + g) + h)(m) = (f + (g + h))(m) for all m M. We have seen from group theory that the zero function 0 : M M, defined by 0(m) = 0 for all m M, is a group homomorphism. Thus 0 End (M). The zero function serves as a neutral element for addition since function addition is commutative and f + 0 = f as (f + 0)(m) = f(m) + 0(m) = f(m) + 0 = f(m) for all m M. Note that f : M M defined by ( f)(m) = f(m) for all m M is a group homomorphism since ( f)(m + n) = (f(m + n)) = (f(m) + f(n)) = ( f(n)) + ( f(m)) = ( f(m)) + ( f(n)) = ( f)(m) + ( f)(n) for all m, n M. The reader is left to show that f is an additive inverse for f. We have finally shown that End (M) is a group under addition. To complete the proof that End (M) is a ring with unity we need to establish the distributive laws. First of all (f +g) h = f h+g h follows by definition of function composition and function addition since ((f + g) h)(m) = (f + g)(h(m)) = f(h(m)) + g(h(m)) = (f h)(m) + (g h)(m) = (f h + g h)(m) for all m M. Since f is a group homomorphism the distributive law f (g + h) = f g + f h holds as (f (g + h))(m) = f((g + h)(m)) = f(g(m) + h(m)) 2
= f(g(m)) + f(h(m)) = (f g)(m) + (f h)(m) = (f g + f h)(m) for all m M. Therefore End (M) is a ring with unity. Comment: The proof actually establishes more. For non-empty sets X, Y let Fun(X, Y ) be the set of all functions f : X Y. Let M be a non-empty set. Then F un(m, M) is a monoid under composition with neutral element I M. Suppose that X is a non-empty set and M is an additive (not necessarily abelian) group. Then F un(x, M), in particular F un(m, M), is a group under function addition with neutral element the zero map 0 : X M defined by 0(x) = 0 for all x X. Furthermore the distributive law holds for all f, g, h, F un(m, M). (f + g) h = f h + g h Let f F un(m, M) be fixed. Then the distributive law f (g + h) = f g + f h holds for all g, h F un(m, M) if and only if f End (M). (To see this let m, n M and g(x) = m and h(x) = n for all x M.) Observe that End (M) is a submonoid of F un(m, M) with neutral element I M. When M is abelian End (M) is a subgroup of F un(m, M) under function addition. (In this case End (M) is a ring with unity under function addition and composition.) Note that I M + I M End (M) if and only if M is abelian. Thus End (M) is closed under function addition if and only if M is abelian. Now suppose that M is a left R-module. (b) (8) For r R define σ r : M M by σ r (m) = r m for all m M. Show that σ r End(M) for all r R and π : R End(M) defined by π(r) = σ r for all r R is a homomorphism of rings with unity. Solution: Let r R. for m, n M the calculation σ r (m + n) = r (m + n) = r m + r n = σ r (m) + σ r (n) shows that σ r : M M is an endomorphism of (additive) groups. 3
Let r, r R. We have just shown that π(r) = σ r End (M). Note that π(r)(m) = σ r (m) = r m for all m M. Since π(r + r )(m) = (r + r ) m = r m + r m = π(r)(m) + π(r )(m) = (π(r) + π(r ))(m) for all m M it follows that π(r + r ) = π(r) + π(r ). Likewise π(rr )(m) = (rr ) m = r (r m) = π(r)(r m) = π(r)(π(r )(m)) = (π(r) π(r ))(m) for all m M shows that π(rr ) = π(r) π(r ). Thus π is a ring homomorphism. Since π(1)(m) = 1 m = m = I M (m) for all m M we have π(1) = I M. Therefore π is a homomorphism of rings with unity. 2. (20 total) Let M be a left R-module. For a non-empty subset S of M the subset of R defined by ann R (S) = {r R r s = 0 s S} is called the annihilator of S. If S = {s} is a singleton we write ann R (s) for ann R ({s}). (a) (8) Suppose that N is a submodule of M. Show that ann R (N) is an ideal of R. Solution: Let I = ann R (N). Then 0 I since 0 m = 0 for all m N. Thus I. Suppose r, r I and n N. Then (r r ) n = r n r n = 0 0 = 0 since n, n N. Thus r r I which establishes that I is an additive subgroup of R. For r R the calculations and (r r) n = r (r n) = r 0 = 0 (rr ) n = r (r n) r N = (0) show that r r, rr I. Therefore I is an ideal of R. 4
Now suppose m M is fixed. (b) (6) Show that ann R (m) is a left ideal of R. Solution: The calculations of part (a) establish p[art (b). (c) (6) Let f : R R m be defined by f(r) = r m for all r R. Show f is a homomorphism of left R-modules and F : R/ann R (m) R m given by F (r + ann R (m)) = r m for all r R is a well-defined isomorphism of left R-modules. Solution: Let r, r R. Then R m is a submodule of M (a proof really is in order) and the calculations and f(r + r ) = (r + r ) m = r m + r m = f(r) + f(r ) f(rr ) = (rr ) m = r (r m) = r f(r ) show that f is a map of left R-modules. One could appeal to the Isomorphism Theorems for R-modules to complete the problem; we will follow the intent of the instructions. F is well-defined. Suppose that r, r R and r + ann R (m) = r + ann R (m). Then r r ann R (m) which means (r r ) m = 0 or equivalently r m = r m. Therefore F (r + ann R (m)) = r m = r m = F (r + ann R (m)) which means F is well-defined. Note that F and f are related by F (r + ann R (m)) = f(r) for all r R. F is a module map since and F ((r + ann R (m)) + (r + ann R (m))) = F ((r + r ) + ann R (m)) = f(r + r ) = f(r) + f(r ) = F (r + ann R (m)) + F (r + ann R (m)) F (r (r + ann R (m))) 5
= F (rr + ann R (m)) = f(rr ) = r f(r ) = r F (r + ann R (m)) for all r, r R. F is surjective since f is. Since Ker F = {r + ann R (m)) r ann R (m))} is the trivial subgroup of R/ann R (m), it follows that the (group) homomorphism F is injective. 3. (20 total) Let k be a field, V a vector space over k, and T End k (V ) be a linear endomorphism of V. Then the ring homomorphism π : k[x] End k (V ) defined by π(f(x)) = f(t ) for all f(x) k[x] determines a left k[x]-module structure on V by f(x) v = π(f(x))(v) = p(t )(v) for all v V. (a) (15) Let W be a non-empty subset of V. Show that W is a k[x]- submodule of V if and only if W is a T -invariant subspace of V. Solution: Suppose that f(x) = α 0 + + α n X n k[x]. Then f(x) v = f(t )(v) = (α 0 I V + + α n T n )(v) = α 0 v + + α n T n (v) for all v V. Let W be a k[x]-submodule. Then W is an additive subgroup of V by definition. Let w W. Since f(x) w = α 0 w when f(x) = α 0 and f(x) w = T (w) when f(x) = X, α 0 w W for all α 0 k, which means that W is a subspace of V, and T (w) W, which means that W is T -invariant (or T -stable). Conversely, let W be a T -invariant subspace of V. Then T m (W ) W for all m 0 by induction on m. Therefore f(x) w W for all w W which means that W is a k[x]-submodule of V. (b) (5) Suppose that V = k[x] v is a cyclic k[x]-module. Show that ann k[x] (V ) = (f(x)), where f(x) is the minimal polynomial of T. There are various ways of defining the minimal polyno- One is the unique monic generator of the ideal I of all Solution: mial of T. 6
f(x) k[x] such that f(t ) = 0 when I (0). Otherwise the minimal polynomial is set to 0 when I = (0). Note that I = ann k[x] (V ). Comment: The condition V is cyclic is not necessary; it was there anticipating a certain application. 4. (20 total) Let M be a left R-module. (a) (5) Suppose that N is a non-empty family of submodules of M. Show that L = N N N is a submodule of M. Solution: Since submodules are (additive) subgroups, we know from group theory that L = N N N is a subgroup of M. Let r R and n L. To complete the proof that L is a submodule of M we need only show that r n L. Since n L, n N for all N N. Hence r n N for all N N, since each N is a submodule of M, and therefore r n L. Since M is submodule of M, it follows that any S subset of M is contained in a smallest submodule of M, namely the intersection of all submodule containing S. This submodule is denoted by (S) and is called the submodule of M generated by S. (b) (5) Let = S M. Show that (S) = {r 1 s 1 + + r l s l l 1, r 1,..., r l R, s 1,..., s l S}. solution: Let L = {r 1 s 1 + + r l s l l 1, r 1,..., r l R, s 1,..., s l S}. Informally we may describe L as the set of all finite sums of products r s, where r R and s S. Now L (S). For since S (S) and (S) is a submodule of M, products r s (S) since (S) is closed under module multiplication, and thus r 1 s 1 + + r l s l (S), by induction on l, for all r 1,..., r l R and s 1,..., s l S since (S) is closed under addition. 7
To complete the proof we need only show (S) L. Since s = 1 s for all s M it follows that S L. Thus to show (S) L we need only show that L is a submodule of M. Since S and S L it follows that L. Suppose that x, y L. Then x, y are finite sums of products r s, where r R and s S; therefore x + y is as well. We have shown x + y L. Since (r s) = ( r) s and r (r s) = (r r) s for r, r R and s S, it follows that x and r x are finite sums of products r s, where r R and s S. Therefore x, r x L which completes our proof that L is a submodule of M. Comment: Here are the highlights of a proof of the fact the L is a submodule of M which follows the literal description of L. Let x, y L. Write x = r 1 s 1 + + r l s l and y = r 1 s 1 + + r l s l, where l, l 1, r 1,..., r l, r 1,..., r l R, and s 1,..., s l, s 1,..., s l S. Thus x + y = r 1 s 1 + + r l s l + r 1 s 1 + + r l s l which means where l = l + l, and x + y = r 1 s 1 + + r l s l, r i = s i = Thus x + y L. Note that and { ri : 1 i l r i l : l < i l + l, { si : 1 i l s i l : l < i l + l. x = (r 1 s 1 ) (r l s l ) = ( r 1 ) s 1 + + ( r l ) s l L r x = r (r 1 s 1 ) + + r (r l s l ) = (rr 1 ) s 1 + + (rr l ) s l L. Suppose f, f : M M are R-module homomorphisms. 8
(c) (5) Show that N = {m M f(m) = f (m)} is a submodule of M. Solution: First of all 0 N since f(0) = 0 = f (0) as f, f are group homomorphisms. Suppose that m, n M. Then f(m n) = f(m + ( n)) = f(m) + f( n) = f(m) f(n). Thus for m, n N we have f(m n) = f(m) f(n) = f (m) f (n) = f (m n) which means m n N. Therefore N M. For r R the calculation f(r m) = r f(m) = r f (m) = f (r m) shows that r m N. Therefore N is a submodule of M. (d) (5) Suppose that S generates M. f(s) = f (s) for all s S. Show that f = f if and only if Solution: If f = f then f(s) = f (s) for all s M, hence for all s S. Conversely, suppose that f(s) = f (s) for all s S and let N be as in part (a). Then S N which means M = (S) N since S generates M and N is a submodule of M. Therefore M = N which means f(m) = f (m) for all m M, or equivalently f = f. Comment: There is no need to invoke part (b) for part (d). 5. (20 total) Use Corollary 2 of Section 2.3 Supplement and the equation of Problem 3 of Written Homework 3 to prove the following: Theorem 1 Let k be a field and suppose that G is a finite subgroup of k. Then G is cyclic. Solution: A proof is to be based on the equations for all positive integers n and ϕ(d) = n d n d G n d ϕ(d) = G 9
for all finite groups G. Suppose that H k is cyclic of order d. Then a d = 1, or equivalently a is a root of X d 1 k[x], for all a H. This polynomial has at most d roots in k since k is a field. Therefore H is the set of the roots of X d 1 in k. We have shown that there is at most one cyclic subgroup of order d in k. Now let G k be finite. We have shown n d = 0 or n d = 1 for each positive divisor of G. Since ϕ(d) > 0 for all positive integers d, from the equations n d ϕ(d) = G = ϕ(d) = 1ϕ(d) d G d G d G we deduce that n d = 1 for all positive divisors d of G. In particular n G = 1 which means that G has a cyclic subgroup of order G ; thus G is cyclic. 10